Do I calculate the integral at every n while $n\rightarrow \infty$, say if $f(x) \in C_b$ (e.g. $f \equiv 1)$ I get always $1$ but at some $n$ I get $0$ since $f(x)$ is bounded and, hence, it does not converge weakly ?

$\begingroup$thanks for the quick reply, if take $\delta_{\frac{1}{n}}$ then I get the weak convergence since $\int f(x)\, d\delta_{\frac{1}{n}} \stackrel{n\rightarrow\infty}{\to} \int f(x)\, \delta_{0}(dx)$ right ?$\endgroup$
– jedMay 3 '13 at 16:11

$\begingroup$Let $\mu_n = N(0, \frac{1}{n})$ then I have $\frac{\sqrt{n}}{\sqrt{2\pi}} \int f(x)\,e^{-\frac{x²n}{2}} dx$ as $n \rightarrow\infty$ the integral goes to $0$. Does this proof that it does not converge weakly ?$\endgroup$
– jedMay 3 '13 at 17:44